WAEC SSCE - Further Mathematics - 2024 (Essay)

Antwoorddetails

From the diagram above, using cosine's law

Cos \(\theta\) = \(\frac{10^2 + 12^2 - 15^2}{2 \times 10 \times 12}\)

Cos \(\theta\) = \(\frac{100 + 144 - 225}{2 \times 10 \times 12}\) = \(\frac{244 - 225}{240}\)

= \(\frac{19}{240}\) = 0.0791667

\(\theta\) = Cos\(^{-1}\)(0.0791667) = 85.459º

But the angle between the two vectors = 180º - \(\theta\) = 180º - 85.459º = 94.54º

Antwoorddetails

Given:  y = 8x + \(\frac{27}{2x^2}\) =  y = 8x + \(\frac{27}{2}\)x\(^{-2}\),

(a) \(\frac{dy}{dx}\) = 8x\(^0\) +  \(\frac{27}{2}\)(-2)x\(^{-3}\)

= 8 - 27x\(^{-3}\) = 8 -  \(\frac{27}{x^3}\)

(b) \(\frac{dy}{dx}\) = 8 -  \(\frac{27}{x^3}\) = 0

= 8 = \(\frac{27}{x^3}\)

= 8x\(^3\) = 27

x\(^3\) = \(\frac{27}{8}\)

x = \( \sqrt[3]{\frac{27}{8}}\) = \(\frac{3}{2}\)

x = \(\frac{3}{2}\)

y = 8x + \(\frac{27}{2x^2}\) = 8(\(\frac{3}{2}\)) + \(\frac{27}{2}\)(\(\frac{3}{2}\))\(^{-2}\)

y = 4(3) +  \(\frac{27}{2}\)(\(\frac{4}{9}\))

y = 12 + 6 = 18

Thus, stationary points are (\(\frac{3}{2}\), 18)

\(\frac{d^2y}{dx^2}\) = - \(\frac{27}{x^3}\) = - 27(-3)x\(^{-4}\) = \(\frac{81}{x^4}\) = \(\frac{81}{(\frac{3}{2})^4}\) = 81 x \(\frac{16}{81}\) = 16 > 0

For all x, the stationary point is a minimum.

(c) gradient = 8 -  \(\frac{27}{x^3}\) = 8 -  \(\frac{27}{2^3}\) = 8 -  \(\frac{27}{8}\) = \(\frac{37}{8}\)

Equation of normal, at (2, 2)

but m\(_2\) =  \(\frac{- 1}{m_1}\) = \(\frac{- 1}{\frac{37}{8}}\) = \(\frac{- 8}{37}\)

y - y\(_1\) = m\(_2\)[x - x\(_1\)]

y - 2 = \(\frac{- 8}{37}\)[x - 2]

37y - 74 = - 8x + 16 

37y + 8x - 90 = 0

Antwoorddetails

From the diagram above;

fr + mg sin \(\theta\) = 420

fr = 420 - 40 x 10 sin 30º

fr = 420 - 200 = 220N

(b) coefficient of friction(μ)  = \(\frac{\text{frictional force}}{\text{normal reaction}}\) = \(\frac{fr}{R}\)

μ =  \(\frac{fr}{mg cos \theta}\) =  \(\frac{220}{400 cos 30}\) = 0.63508

Therefore, μ = 0.635.

Antwoorddetails

\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{A}{2x + 1}\) + \(\frac{Bx + C}{x^2 + 1}\) = \(\frac{A(x^2 + 1) + (Bx + C)(x + 1)}{(2x + 1)(x^2 + 1)}\) - - - - -- - - -(i)

Using the cover-up method, put x = \(\frac{-1}{2}\)

A = \(\frac{9x}{x^2 + 1}\) = \(\frac{9(\frac{-1}{2})}{(\frac{-1}{2})^2 + 1}\) = \(\frac{\frac{-9}{2}}{\frac{5}{4}}\) = \(\frac{-18}{5}\)

From equation (i)

9x = A(\(x^2\) + 1) + (Bx + C)(x + 1)

9x = A\(x^2\) + A  + 2B\(x^2\) + Bx + 2x + C

9x = (A + B)\(x^2\) + (B + 2C)x + (A + C)

comparing coefficients(x\(^2\)

A + 2B = 0

but A =  \(\frac{-18}{5}\)

- A = 2B

-(\(\frac{-18}{5}\)) = 2B

18 = 10B

B =  \(\frac{18}{10}\) =  \(\frac{9}{5}\).

comparing coefficients (constant terms)

A + C = 0

C = - A = -(\(\frac{-18}{5}\))

C = \(\frac{18}{5}\)

\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{\frac{-18}{5}}{2x + 1}\) + \(\frac{\frac{9x}{5} + \frac{18}{5}}{x^2 + 1}\) 

= \(\frac{-18}{5(2x + 1)}\) + \(\frac{9x}{5(x^2 + 1)}\) + \(\frac{18}{5(x^2 + 1)}\)

= \(\frac{18}{5}\)[\(\frac{1}{x^2 + 1}\) + \(\frac{1}{2(x^2 + 1)}\) - \(\frac{1}{2x + 1}\)]

2(b) \(^{2m}P_2\) - 10 = \(^m P_2\)

\(\frac{(2m)!}{(2m - 2)!}\) - 10 = \(\frac{m!}{(m - 2)!}\)

\(\frac{(2m)(2m - 1) \times (2m - 2)!}{(2m - 2)!}\) - 10 = \(\frac{m \times (m - 1) \times (m - 2)!}{(m - 2)!}\)

⇒ 2m(2m - 1) - 10 = m(m - 1)

4m\(^2\) - 2m - 10 = m\(^2\) - m 

4m\(^2\) -  m\(^2\) - 2m + m - 10 = 0

3 m\(^2\) - m - 10 = 0

(3m + 5)(m - 2) = 0

m = 2( +ve value only)

Antwoorddetails

Given:  11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). median = 13\(\frac{1}{2}\) = \(\frac{27}{2}\)

Median = \(\frac{(2x + y) + (x + 2y)}{2}\) = \(\frac{27}{2}\)

\(\frac{3x + 3y}{2}\) = \(\frac{27}{2}\)

3x + 3y = 27

x + y = 9  - - - - -  - - -(i)

also, y = x + 1  - - - - -(ii)

put y = x + 1 into eqn (ii)

x + x + 1 = 9

2x = 9 - 1 = 8

x = \(\frac{8}{2}\) = 4

y = x + 1 = 4 + 1 = 5

(b) Since x = 4 and y = 5

11, 12, (2(3) + 5), (4 + (2(5)), 14, (5\(^2\) - 2(4))

11, 12, 13, 14, 14, 17

Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) = \(\frac{11 + 12 + 13 + 14 + 14 + 17}{6}\) = 13.5 

                                                                                                                 \(\sum f(x - \overline{x})^2\) = 21.50

S.D = \(\sqrt{\frac{\sum f(x - \overline{x})^2}{\sum {f}}}\) = \(\sqrt{\frac{21.50}{6}}\) = 1.89

Antwoorddetails

(NOTE): Frequency density = \(\frac{\text{frequency}}{\text{Class size}}\)

SEE THE DIAGRAM ABOVE

Antwoorddetails

70% passed = p

30% failed = q

Using binomial theorem

P[x = x] = \(\begin{pmatrix} n \\ x \end{pmatrix}\)P\(^x\)q\(^{n - x}\)

(a) Probability that at least two of them failed:

P(X \(\geq\) 2) = 1 − P(X=0) − P(X=1)

When P(x = 0) = \(\begin{pmatrix} 12 \\ 12 \end{pmatrix}\)\(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^2\)\(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^0\)

= 0.01384129

When P(x = 1) = \(\begin{pmatrix} 12 \\ 11 \end{pmatrix}\)\(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^{11}\)\(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^1\)

= 0.07118376

P(X \(\geq\) 2) = 1 - 0.01384129 - 0.0711837 = 0.9149751 ≈ 0.915

(b) half of twelve = 6

P(x = 6) =  \(\begin{pmatrix} 12 \\ 6 \end{pmatrix}\) \(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^6\) \(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^6\)

= 924 x 0.117649 x 0.000729 = 0.079245

(c) \(\frac{1}{6}\) x 12 = 2, i.e. P[x \(\leq\) 2) = P[x = 0] + P[ x = 1] + P[x = 2]

= 0.01384129 + 0.07118376 +  \(\begin{pmatrix} 12 \\ 10 \end{pmatrix}\) \(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^{10}\) \(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^2\)

= 0.01384129 + 0.07118376 + 0.167790298 = 0.252815348.

Antwoorddetails

Given: 

P: (x, y) → (-4x - y, 2x)

Q: (x, y) → (y, 6x - 9y)

R: (x, y) → (x - 2y, 3x + 5y)

(a) the matrix of P, Q, and R

P = \(\begin{pmatrix} - 4  & -1 \\ 2 & 0 \end{pmatrix}\), Q =  \(\begin{pmatrix} 0  & 1 \\ 6 & -9 \end{pmatrix}\), R =  \(\begin{pmatrix} 1  & -2 \\ 3 & 5 \end{pmatrix}\),

(b) (i) 2P - 3R + Q

= 2\(\begin{pmatrix} - 4  & -1 \\ 2 & 0 \end{pmatrix}\) - 3\(\begin{pmatrix} 1  & -2 \\ 3 & 5 \end{pmatrix}\) +  \(\begin{pmatrix} 0  & 1 \\ 6 & -9 \end{pmatrix}\)

=  \(\begin{pmatrix} -8  & -2 \\ 4 & 0 \end{pmatrix}\) -  \(\begin{pmatrix} 3  & -6 \\ 9 & 15 \end{pmatrix}\) +  \(\begin{pmatrix} 0  & 1 \\ 6 & -9 \end{pmatrix}\)

=  \(\begin{pmatrix} -11  & 5 \\ 1 & -24 \end{pmatrix}\)

(ii) QR = \(\begin{pmatrix} 0  & 1 \\ 6 & -9 \end{pmatrix}\)\(\begin{pmatrix} 1  & -2 \\ 3 & 5 \end{pmatrix}\)

 = \(\begin{pmatrix} 0 + 3  & 0+ 5 \\ 6 - 27 & -12 - 45 \end{pmatrix}\)

= \(\begin{pmatrix} 3  & 5 \\ -21 & -57 \end{pmatrix}\)

(iii) the inverse of the matrix R. \(\begin{pmatrix} 1  & -2 \\ 3 & 5 \end{pmatrix}\),

|R| = (1 x 5) - ( -2 x 3) = 5 + 6 = 11

R\(^{-1}\) = \(\frac{Adj (R)}{|R|}\) = \(\frac{1}{11}\)\(\begin{pmatrix} 5  & 2 \\ -3 & 1 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{5}{11}  & \frac{2}{11} \\ \frac{-3}{11} & \frac{1}{11} \end{pmatrix}\)

Antwoorddetails

hyp\(^2\) = 3\(^2\) + 1\(^2\) (Pythagoras's theorem)

hyp = \(\sqrt{10}\)

Given: tan x =  \(\frac{1}{3}\) ( sin and cos will be in the third quadrant and are both negative where 180º < x < 270º)

sin x =  \(\frac{-1}{\sqrt{10}}\), cos x = \(\frac{-3}{\sqrt{10}}\)

\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) =  \(\frac{2sinx cos x - cos x}{2 tan x + 2sin x cos x}\) but 2snx cox = \(\frac{3}{5}\)

\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{2(\frac{1}{3}) + \frac{3}{5}}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{\frac{19}{15}}\)

= (\(\frac{3}{5}\) + \(\frac{3\sqrt{10}}{10}\)) \(\div\) \(\frac{19}{15}\) = \(\frac{15}{19}\)[\(\frac{3}{5}\) + \(\frac{3}{\sqrt{10}}\)]

= \(\frac{9}{19}\) + \(\frac{45}{19\sqrt{10}}\)

= \(\frac{9}{19}\) + \(\frac{45 \sqrt{10}}{19(10)}\)

= \(\frac{9}{19}\) + \(\frac{9 \sqrt{10}}{19(2)}\)

= \(\frac{9}{19}\)[1 + \(\frac{\sqrt{10}}{2}\)]

Antwoorddetails

(a)i  X = \(\frac{1}{5}\)t\(^2\) +  \(\frac{1}{20}\)t\(^3\) ( after dividing thru by 20)

\(\frac{dX}{dt}\) = velocity = \(\frac{2}{5}\)t + \(\frac{3}{20}\))t\(^2\)

at t = 15, v = \(\frac{2}{5}\)(15) + \(\frac{3}{20}\)(15)\(^2\) = 6 + 33.75 = 39.75m/s

Acceleration = \(\frac{dV}{dt}\) =  \(\frac{2}{5}\)t + \(\frac{3}{20}\)t\(^2\) =  \(\frac{2}{5}\) + \(\frac{3}{10}\)t

for initial acceleration, t = 0

a =  \(\frac{2}{5}\)ms\(^2\)

(8) \(\frac{2}{5}\) = \(\frac{2}{5}\) + \(\frac{3}{10}\)t

40 + 30t = 320

30t = 320 - 40 = 280

t = \(\frac{280}{30}\) = 9\(\frac{1}{3}\)secs.

(b) m\(_1\)u\(_1\) + m\(_2\)u\(_2\) = v\(_c\)(m\(_1\) + m\(_2\))

m\(_1\) = 6kg, u\(_1\) = 7m/s, m\(_2\) = ?, u\(_2\) = - 5 m/s

6(7) + m\(_2\)(-5) = 4(6 + m\(_2\))

42 - 5m\(_2\) = 24 + 4m\(_2\)

42 - 24 =  4m\(_2\) + 5m\(_2\)

9m\(_2\) = 18

m\(_2\) = \(\frac{18}{2}\) = 2 kg.

Antwoorddetails

W = 12 N

From the diagram above,

\(\sum{F_x}\) = - 3.938 N, \(\sum{F_y}\) = 3.108 N

Expressing in column vector

\(\begin{pmatrix} i & j \\ 10cos 90º & 10sin90º \\ 16cos180º & 16sin180º \\ 7cos30º & 7sin30º \\ 12cos 300º & 12sin300º \end{pmatrix}\)

Resultant R = \(\sqrt{(F_x)^2 + (F_y)^2}\)

R = \(\sqrt{( - 3.938)^2 + (3.108)^2}\) = 5.018N ≈ 5.02 N

(ii) R = ma (from Newton's law)

But, W = mg 

m = \(\frac{\text{W}}{\text{g}}\) = \(\frac{12}{10}\) = 1.2 kg

From, R = ma, then, a = \(\frac{\text{R}}{\text{m}}\) = \(\frac{5.018}{1.2}\) = 4.182ms\(^{-2}\) ≈ 4.18 ms\(^{-2}\)

Antwoorddetails

Given: 403, 404, 405, 406, . . . . 602

Numbers divisible by 7 are 406, 413, 420, . . . . . 602.

S\(_n\) = \(\frac{n}{2}\)[2a + (n - 1)d]

a = 406, d = 7

T\(_n\) = a + (n - 1)d

602 = 406 + (n - 1)7

602 - 406 = 7n - 7

7n = 602 - 406 + 7 = 203

n = \(\frac{203}{7}\) = 29.

S\(_29\) = \(\frac{29}{2}\)[2(406) + (29 - 1)7]

= \(\frac{29}{2}\)[812+ (28 x 7)]

= \(\frac{29}{2}\)[812+ (196)]

= \(\frac{29}{2}\)[1008] = 14616